Prove that $u-i$ is a maximal ideal of $\frac {\Bbb C[X,Y]} {\left (X^2+Y^2-1 \right )}.$

Question

Let $A = \frac {\Bbb C[X,Y]} {\left (X^2+Y^2-1 \right )}.$ Let $u=X+iY.$ Show that $(u-i)$ is a maximal ideal of $A.$

Any ideal of $A$ is an ideal of $\Bbb C[X,Y]$ containing the ideal $\left ( X^2+Y^2-1 \right ).$ Need only to show that $(u-i)$ is maximal with respect to all the ideals of $\Bbb C[X,Y]$ containing $\left ( X^2+Y^2-1 \right ).$ It is easy to show that $X^2+Y^2-1 \in (u-i)$ since $$\left ( X +i \left (Y-1 \right ) \right ) \cdot \frac {\left (X^2+Y^2-1 \right ) \left (X-i \left (Y-1 \right ) \right )} {X^2+ \left (Y-1 \right )^2} = X^2+Y^2-1.$$ But how do I prove that $(u-i)$ is maximal with respect to all the ideals of $\Bbb C[X,Y]$ containing $\left (X^2+Y^2-1 \right )$?

Any help will be highly appreciated. Thank you very much.

Answer 1

$(X+iY-i)$ is a maximal ideal of $A$ iff $I=(X+iY-i,X^2+Y^2-1)$ is a maximal ideal of $\Bbb C[X,Y]$.

Now, $$(X-i(Y-1))(X+i(Y-1))=X^2+Y^2-2Y+1\in I$$ and so $$Y-1=\frac{X^2+Y^2-1-(X^2+Y^2-2Y+1)}2\in I.$$ Then $$X=X+iY-1-i(Y-1)\in I.$$ So $J\subseteq I$ where $J=(X,Y-1)$. But as $X+iY-Y=X+i(Y-1)$ and $X^2+Y^2-1=XX+(Y+1)(Y-1)$, then $I\subseteq J$ and $I=J$.

But $J$ is the kernel of the $\Bbb C$-algebra homomorphism $\phi: \Bbb C[X,Y]\to\Bbb C$ taking $X$ and $Y$ to $0$ and $1$ respectively. As $\phi$ is a surjective homomorphism to a field, its kernel is a maximal ideal.